Denote by \(\Phi (z) = H(\arg z) \mathrm {Im}(z)^{d-2}\)  the density of \(\mathbf {E}[\mu ^0]\)  with respect to Lebesgue measure. As above, let \(f_t(z) = X_t^z + i Y_t^z\) . Let \(\Upsilon _t^z = Y_t^z/|f_t^{\prime }(z)|\)  and note that \(\Upsilon _t^z\)  is half the conformal radius of \(z\)  in \(\mathbf {H}\setminus \eta ([0,t])\) , which is decreasing in \(t\) . Let \(\Theta _t^z = \arg f_t(z)\) . By (REF ) we have for all \(z \in \mathbf {H}\)  that
\(\mathbf {E}[ \mu ^0(dz) \,|\,\eta |_{[0,t]} ] = |f_t^{\prime }(z)|^{2-d} \Phi (f_t(z)) dz = H(\Theta _t^z) (\Upsilon _t^z)^{d-2} dz.\) 

We note that the filtration generated by \(\eta \)  is the same as the filtration generated by the driving Brownian motion. In particular, the right hand side above is a martingale adapted to a filtration generated by a Brownian motion and therefore has a continuous modification. We shall assume that we are working with this modification. Since the process \((\Upsilon _t^z)^{d-2}\)  is continuous in \(t\)  and positive (up until \(z\)  is swallowed) it follows that \(t \mapsto H(\Theta _t^z)\)  is also continuous (up until \(z\)  is swallowed).
We will argue that \(H|_{[\arg z,\pi )}\)  has a continuous version; the same argument implies that \(H|_{(0,\arg z]}\)  has a continuous version (and the two versions agree at \(\arg z\)  so \(H\)  has a continuous version). Let \(\tau \)  be the time that \(z\)  is swallowed and assume that we are working on the event that \(\Theta _\tau ^z = \pi \) . We first note that there are a.s. only countably many times \(t_0\)  so that \(\Theta ^z\)  has a local minimum or maximum at the time \(t_0\) . Then as \(\Theta ^z\)  a.s. hits every \(\theta \in [\arg z,\pi )\)  uncountably many times ([1]}, using that \(\Theta ^z\)  is up to time change locally absolutely continuous with respect to a standard Brownian motion) it is a.s. the case that for every \(\theta \in [\arg z,\pi )\)  there is a time \(t_\theta \)  so that \(\Theta _{t_\theta }^z = \theta \)  and \(\Theta ^z\)  does not have a local minimum or maximum at the time \(t_\theta \) . Let \(t_\theta ^{\prime }\)  be any other time so that \(\Theta _{t_\theta ^{\prime }}^z = \theta \) . We claim that \(H_{t_\theta } = H_{t_\theta ^{\prime }}\) . To see this, we first note that it is a.s. the case that \(\Theta ^z\)  is not constant on any open interval of time. Thus by the continuity of \(\Theta ^z\)  and since \(H_t = H(\Theta _t^z)\)  for a.e. \(t \in [0,\tau ]\)  we can find sequences \((a_n)\) , \((a_n^{\prime })\)  so that \(a_n \rightarrow t_\theta \)  and \(a_n^{\prime } \rightarrow t_\theta ^{\prime }\) , \(\Theta _{a_n}^z = \Theta _{a_n^{\prime }}^z\)  for all \(n \in \mathbf {N}\) , and \(H_{a_n} = H(\Theta _{a_n}^z)\) , \(H_{a_n^{\prime }} = H(\Theta _{a_n^{\prime }}^z)\)  for all \(n \in \mathbf {N}\) . That is, \(H_{a_n} = H_{a_n^{\prime }}\)  for all \(n \in \mathbf {N}\)  hence \(H_{t_\theta } = H_{t_\theta ^{\prime }}\) . For \(\theta \in [\arg z,\pi )\)  we then define \(\widetilde{H}(\theta ) = H_{t_\theta }\) . Then we have that \(\widetilde{H}(\Theta _t^z) = H_t\)  for all \(t \ge 0\)  and, arguing as above, it is not difficult to see that \(\widetilde{H}\)  is continuous. We also note that \(\widetilde{H}\)  is not random. Indeed, suppose that \((H_t^1, \Theta _t^{z,1})\) , \((H_t^2, \Theta _t^{z,2})\)  are independent copies of \((H_t, \Theta _t^z)\)  conditioned on \(\Theta _t^{z,i}\)  hitting \(\pi \)  before 0. Then if we define \(t_\theta ^i\)  for \((H_t^i, \Theta _t^{z,i})\) , \(i=1,2\) , in the same way that we defined \(t_\theta \)  for \((H_t, \Theta _t^z)\)  we a.s. have that \(H_{t_\theta ^1}^1 = H_{t_\theta ^2}^2\)  for all \(\theta \in [\arg z,\pi )\) . That is, the function \(\widetilde{H}\)  produced from \((H_t^1, \Theta _t^{z,1})\)  a.s. agrees with the function produced from \((H_t^2, \Theta _t^{z,2})\) . It is left to show that \(\widetilde{H}\)  is a version of \(H |_{[\arg z,\pi )}\) , i.e., \(\widetilde{H}(\theta ) = H(\theta )\)  for a.e.  \(\theta \in [\arg z,\pi )\) . We note that \(H_1 = \widetilde{H}(\Theta _1^z)\)  is equal to \(H(\Theta _1^z)\)  with probability 1. As the law of \(\Theta _1^z\)  has a positive density on \([\arg z,\pi )\)  with respect to Lebesgue measure, we conclude that \(\widetilde{H}\)  is a version of \(H|_{[\arg z,\pi )}\) .
Since the measure defined by the density \(H(\arg z) \mathrm {Im}(z)^{d-2}\)  with respect to Lebesgue measure is not affected by changing \(H\)  on a set of Lebesgue measure zero we may thus assume that \(H\)  is continuous. Fix some \(\delta > 0\)  and let \(\tau _z = \tau _z^\delta = \inf \lbrace t \ge 0: \Upsilon _t^z \le \delta \rbrace \) . Fix \(r, t > 0\)  and let \(\sigma _z = \tau _z \wedge \inf \lbrace s \ge 0: H(\Theta _s^z) \ge r\rbrace \) . Note that this is a stopping time by the continuity of \(s \mapsto H(\Theta _s^z)\) . Let also \(\sigma _z(t) = t \wedge \sigma _t\) .
Taking expectations of (REF ) at the time \(\sigma _z(t)\)  we have that
\(\Phi (z) = \mathbf {E}[ H(\Theta _{\sigma _z(t)}^z) (\Upsilon _{\sigma _z(t)}^z)^{d-2} ].\) 

Next, we consider the continuous local martingale [2]}
\(M_t^z = (Y_t^z)^{d-2} \sin (\Theta _t^z)^{\frac{8}{\kappa }-1} |f_t^{\prime }(z)|^{2-d} = (\Upsilon _t^z)^{d-2} \sin (\Theta _t^z)^{\frac{8}{\kappa }-1}.\) 

Let \(\mathbf {P}_z^*\)  denote the measure obtained by weighting the underlying probability measure \(\mathbf {P}\)  by \(M_t^z\) . By [2]}, under the measure \(\mathbf {P}_z^*\) , \(\eta \)  has the law of an \(\mathrm {SLE}_\kappa (\kappa -8)\)  process in \(\mathbf {H}\)  with force point at \(z\) . Note that if we denote by \(\mathbf {E}_z^*\)  the expectation with respect to \(\mathbf {P}_z^*\) , then (REF ) implies that
\(\Phi (z) &= \mathbf {E}\!\left[ H(\Theta _{\sigma _z(t)}^z) (\Upsilon _{\sigma _z(t)}^z)^{d-2}\right] = M_0^z \mathbf {E}_z^*\!\left[H(\Theta _{\sigma _z(t)}^z)/\sin (\Theta _{\sigma _z(t)}^z)^{\frac{8}{\kappa }-1}\right].\) 

Next, we shall prove that \(\Phi \)  is a smooth function.
Let \(a = 2/\kappa \) . We parameterize \(\eta \)  so that
\(\partial _t g_t(z) = \frac{a}{g_t(z) - B_t}, \quad g_0(z) = z.\) 

Then under the measure \(\mathbf {P}_z^*\) , \(\Theta _t^z\)  satisfies
\(d \Theta _t^z = \frac{2a X_t^z Y_t^z}{((X_t^z)^2 + (Y_t^z)^2)^2} dt + \frac{Y_t^z}{(X_t^z)^2+(Y_t^z)^2} dW_t,\) 

where \(W\)  is a \(\mathbf {P}_z^*\) -Brownian motion. Let \(\widetilde{t}(s)\)  be the solution to
\(s = \int _0^{\widetilde{t}(s)} \frac{(Y_u^z)^2}{((X_u^z)^2 + (Y_u^z)^2)^2} du.\) 

Then \(\widetilde{\Upsilon }_s^z = \Upsilon _{\widetilde{t}(s)}^z\)  satisfies \(d \widetilde{\Upsilon }_s^z = -2a \widetilde{\Upsilon }_s^z ds\) , that is, \(\widetilde{\Upsilon }_s^z = \widetilde{\Upsilon }_0^z e^{-2as} = \mathrm {Im}(z) e^{-2as}\) . Moreover, letting \(\widetilde{\Theta }_s^z = \Theta _{\widetilde{t}(s)}^z\)  we have that \(\widetilde{\Theta }_s^z\)  under \(\mathbf {P}_z^*\)  satisfies
\(d \widetilde{\Theta }_s^z = 2a \cot (\widetilde{\Theta }_s^z)ds + d\widetilde{W}_s, \quad \widetilde{\Theta }_0^z = \arg z,\) 

where \(\widetilde{W}_s\)  is a standard Brownian motion. Thus, under \(\mathbf {P}_z^*\) , \(\widetilde{\Theta }_s^z\)  has the law of a radial Bessel process of dimension \(4a+1\) , and thus, by [4]}, has transition density given by
\(p_s(x,y) = \sin (y) \sum _{n=0}^\infty \frac{(1-\cos ^2(y))^{2a-1/2} C_n^{(2a)}(\cos (x)) C_n^{(2a)}(\cos (y))}{\int _{-1}^1 (1-u^2)^{2a-1/2} C_n^{(2a)}(u)^2 du} \exp \!\left(-\frac{n}{2}(n+4a)t \right),\) 

where \(C_n^{(\alpha )}\)  denotes the Gegenbauer polynomial with degree \(n\)  and index \(\alpha \) , see [5]}. Thus, the function \(q(z) = p_s(\arg z,y)\)  is smooth. Note that \(\rho =\kappa -8 < \kappa /2-2\)  for \(\kappa \in (4,8)\)  so that \(\tau _z^\delta \)  is \(\mathbf {P}_z^*\) -a.s. finite. We assume that \(\delta \in (0,\Upsilon _0^z)\) . Taking a limit as \(t \rightarrow \infty \)  and then \(\delta \rightarrow 0\)  in (REF ) and applying Fatou's lemma we see that
\(\Phi (z)&= \liminf _{\delta \rightarrow 0} \liminf _{t \rightarrow \infty } M_0^z\mathbf {E}_z^*\!\left[H(\Theta _{\sigma _z(t)}^z)/\sin (\Theta _{\sigma _z(t)}^z)^{\frac{8}{\kappa }-1}\right]\\&\ge M_0^z \mathbf {E}_z^*\!\left[\liminf _{\delta \rightarrow 0} \liminf _{t \rightarrow \infty } H(\Theta _{\sigma _z(t)}^z)/\sin (\Theta _{\sigma _z(t)}^z)^{\frac{8}{\kappa }-1}\right].\) 

Since \(\sin (\theta )^{1-8/\kappa } \ge 1\)  for \(\kappa \in (4,8)\)  and \(\theta \in (0,\pi )\)  we have that the right hand side is bounded from below by \(r \wedge \sup _{\theta \in (0,\pi )} H(\theta )\) . By taking a limit as \(r \rightarrow \infty \)  we thus see that \(H\)  is bounded by \(\Phi (z) / M_0^z\) . Altogether, this implies that \(H(\widetilde{\Theta }_s^z) / \sin (\widetilde{\Theta }_s^z)^{\frac{8}{\kappa }-1}\)  is a continuous \(\mathbf {P}_z^*\) -martingale which is in \(L^p\)  for some \(p > 1\)  (as \(\sin (\widetilde{\Theta }_s^z)^{1-\frac{8}{\kappa }}\)  has a finite \(p\) th \(\mathbf {P}_z^*\) -moment for some \(p > 1\)  and we have just argued that \(H\)  is bounded). Thus we can take a limit as \(t \rightarrow \infty \)  and \(r \rightarrow \infty \)  to see that
\( \Phi (z) = M_0^z \mathbf {E}_z^*\!\left[H(\widetilde{\Theta }_s^z)/\sin (\widetilde{\Theta }_s^z)^{\frac{8}{\kappa }-1}\right]\) 
where \(s\)  is such that \(\widetilde{\Upsilon }_s^z = \delta \) ; note that \(s > 0\) . As \(p_s(\arg z,y)\)  is smooth, this implies that \(\Phi (z)\)  is smooth as well. Since \(\Phi \)  and \(\mathrm {Im}(z)\)  are smooth in \(\mathbf {H}\) , so is \(H(z)\) .
Moreover, by (REF ) we have that
\(\Phi (f_t(z)) |f_t^{\prime }(z)|^{2-d} = H(\Theta _t^z) (\Upsilon _t^z)^{d-2}\) 

is a continuous local martingale. By the symmetry of \(\mathrm {SLE}\) , as well as the zero-boundary GFF, it follows that \(H(\tfrac{\pi }{2} + \theta ) = H(\tfrac{\pi }{2} - \theta )\)  for all \(\theta \in (0,\tfrac{\pi }{2})\) . Since \(H\)  is smooth, it follows that \(H^{\prime }(\tfrac{\pi }{2}) = 0\)  and this together with Itô's formula (since \(H(\Theta _t^z) (\Upsilon _t^z)^{d-2}\)  is a local martingale), implies that
\(H^{\prime \prime }(\theta ) + \left( 2 - \frac{8}{\kappa } \right) \cot (\theta )H^{\prime }(\theta ) + \left( \frac{8}{\kappa } - 1 \right) H(\theta ) = 0, \quad H\!\left(\frac{\pi }{2} \right) = c, \quad H^{\prime }\!\left(\frac{\pi }{2} \right) = 0,\) 

for some \(c>0\) . By a standard uniqueness result for linear second-order differential equations, there exist two linearly independent solutions to (REF ), namely \(H \equiv 0\)  and \(H(\theta ) = c \sin ^{\frac{8}{\kappa }-1}(\theta )\) . Thus, since \(H\)  is clearly not 0, the result follows.
